With what angular velocity the earth should spin in order that a body lying at `37^@` latitude may become weightless.

Options

A. `5/4 sqrt(g/R)`
B. `25/16sqrt(g/R)`
C. `5/3sqrt(g/R)`
D. `25/9sqrt(g/R)`

Explanation

To determine the angular velocity at which the Earth should spin for a body lying at 37° latitude to become weightless, we can follow these steps:

### Step 1: Understand the Concept of Weightlessness
A body is said to be weightless when the effective gravitational force acting on it becomes zero. The effective weight (or apparent weight) of the body at latitude θ can be expressed as:
\[ g' = g - \omega^2 r \cos^2 \theta \]
where:
- \( g' \) is the effective acceleration due to gravity,
- \( g \) is the standard acceleration due to gravity,
- \( \omega \) is the angular velocity of the Earth,
- \( r \) is the radius of the Earth,
- \( \theta \) is the latitude.

### Step 2: Set the Effective Weight to Zero
For the body to be weightless, we set \( g' = 0 \):
\[ 0 = g - \omega^2 r \cos^2 \theta \]
Rearranging gives:
\[ \omega^2 r \cos^2 \theta = g \]

### Step 3: Solve for Angular Velocity \( \omega \)
From the equation above, we can solve for \( \omega \):
\[ \omega^2 = \frac{g}{r \cos^2 \theta} \]
Taking the square root:
\[ \omega = \sqrt{\frac{g}{r \cos^2 \theta}} \]

### Step 4: Substitute the Values
We know that for \( \theta = 37^\circ \):
- \( \cos 37^\circ = \frac{4}{5} \)

Substituting this value into the equation:
\[ \omega = \sqrt{\frac{g}{r \left(\frac{4}{5}\right)^2}} \]
This simplifies to:
\[ \omega = \sqrt{\frac{g}{r \cdot \frac{16}{25}}} \]
\[ \omega = \sqrt{\frac{25g}{16r}} \]
\[ \omega = \frac{5}{4} \sqrt{\frac{g}{r}} \]

### Step 5: Final Result
Thus, the angular velocity \( \omega \) at which the Earth should spin for a body at 37° latitude to become weightless is:
\[ \omega = \frac{5}{4} \sqrt{\frac{g}{r}} \]